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In quadrilateral ABCD,ABCD,ABCD, AB=AD=1,AB = AD = 1,AB=AD=1, BD=2,BD = \sqrt{2},BD=2, BC=22,BC = \dfrac{ \sqrt{2} }{ 2 },BC=22, and CD=62.CD = \dfrac{ \sqrt{6} }{ 2}.CD=26. The area of triangle ABCABCABC can be expressed in the form a+bc \dfrac{\sqrt{a} + b}{c} ca+b, where a,a, a, b,b, b, and ccc are positive integers and aaa is square-free. What is the value of a+b+ca+b+ca+b+c?
This problem is posed by Michael T.
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